# Physics 616

• Prof. Andrew W. Steiner
(or Andrew or "Dr. Steiner")
• Office hour: 103 South College, Thursday 11am
• Email: awsteiner@utk.edu
• Homework: Electronically as .pdf
• You may work with each other on the homework, but you must write the solution in your own words

Use down and up arrows to proceed to the next or previous slide.

## Outline

• Transformation of Basis Vectors and Curved Coordinates

## Equivalence Principle

• Weak form: All frames freely falling in a gravitational field are inertial frames (proved in Schutz)
• Strong form: All experiments done performed in a frame freely falling in a gravitational field are equivalent (not necessarily correct, but no evidence to the contrary)
• Strong form applicable to self-gravitating objects
• Requires e.g. G constant everywhere
• Requires that orbital motion in strong self-gravity is the same as that in weak self-gravity

## Tests of the Strong Equivalence Principle

 Hierarchical triple system: one star loosely orbiting a binary system Great for testing the strong equivalence principle PSR B1620-26: white-dwarf, pulsar, and a planet PSR B0337+1715: a white dwarf orbiting a white dwarf-neutron star binary video Ransom et al. (2013)

## Jacobians

• Variable transformation $$\xi = \xi(x,y) \qquad \eta = \eta(x,y)$$ and $$\begin{eqnarray} d \xi &=& \frac{\partial \xi}{\partial x} d x + \frac{\partial \xi}{\partial y} d y \\ d \eta &=& \frac{\partial \eta}{\partial x} d x + \frac{\partial \eta}{\partial y} d y \\ \end{eqnarray}$$
• The relationship between derivatives is just a linear transformation $$\left( \begin{array}{c} d \xi \\ d \eta \end{array} \right) = \left( \begin{array}{cc} \frac{\partial \xi}{\partial x} & \frac{\partial \xi}{\partial y} \\ \frac{\partial \eta}{\partial x} & \frac{\partial \eta}{\partial y} \\ \end{array} \right) \left( \begin{array}{c} d x \\ d y \end{array} \right) = J \left( \begin{array}{c} d x \\ d y \end{array} \right)$$
• Useful in integration $$\int f(\xi,\eta)~d\xi~d\eta = \int f[\xi(x,y),\eta(x,y)]~|\mathrm{det} J|~ d x~d y$$
• What if $|\mathrm{det} J|$ is infinite somewhere?

## Connection to Lorentz Transformations

• We define $${\Lambda^{\alpha^{\prime}}}_{\beta} = \left( \begin{array}{cc} \frac{\partial \xi}{\partial x} & \frac{\partial \xi}{\partial y} \\ \frac{\partial \eta}{\partial x} & \frac{\partial \eta}{\partial y} \\ \end{array} \right)$$ for an arbitrary coordinate transformation (not just a SR boost), remembering that $$V^{\alpha^{\prime}} = {\Lambda^{\alpha^{\prime}}}_{\beta} V^{\beta}$$
• One can also transform derivatives $$\frac{\partial \phi}{\partial \xi} = \frac{\partial x}{\partial \xi} \frac{\partial \phi}{\partial x} + \frac{\partial y}{\partial \xi} \frac{\partial \phi}{\partial y}$$
• Which gives $$\left(\tilde{d}\phi\right)_{\beta^{\prime}} = {\Lambda^{\alpha}}_{\beta^{\prime}} \left(\tilde{d}\phi\right)_{\alpha}$$

## Inverse Transformations

• The inverse of $$J=\left[ \begin{array}{cc} \left(\frac{\partial \xi}{\partial x}\right)_y & \left(\frac{\partial \xi}{\partial y}\right)_x \\ \left(\frac{\partial \eta}{\partial x}\right)_y & \left(\frac{\partial \eta}{\partial y}\right)_x \\ \end{array} \right]$$ is just $$J^{-1} = \left[ \begin{array}{cc} \left(\frac{\partial x}{\partial \xi}\right)_{\eta} & \left(\frac{\partial x}{\partial \eta}\right)_{\xi} \\ \left(\frac{\partial y}{\partial \xi}\right)_{\eta} & \left(\frac{\partial y}{\partial \eta}\right)_{\xi} \\ \end{array} \right]$$
• Keep in mind $$\left(\frac{\partial \xi}{\partial x}\right)_y^{-1} = \left(\frac{\partial x}{\partial \xi}\right)_y \neq \left(\frac{\partial x}{\partial \xi}\right)_{\eta} \, !$$

## Polar Coordinate Basis

• Using the fact that basis vectors transform like one-forms, $$\vec{e}_r = {\Lambda^x}_r \vec{e}_x + {\Lambda^y}_r \vec{e}_y = \frac{\partial x}{\partial r} \vec{e}_x + \frac{\partial y}{\partial r} \vec{e}_y$$ and similarly for $\vec{e}_{\theta}$
• To transform gradients $$\tilde{d} \theta = \frac{\partial \theta}{\partial x} \tilde{d} x + \frac{\partial \theta}{\partial y} \tilde{d} y$$
• Unlike in Cartesian coordinate systems, the basis one-forms do not have unit magnitude $$\left| \vec{e}_{\theta} \right|^{2} = r^2$$

## Norms in the curved space

• The metric tensor, again, tells us how to define four-vector lengths $$\vec{U} \cdot \vec{V} = g_{\alpha \beta} V^{\alpha} V^{\beta}$$ where $$g_{\alpha \beta} = \left( \begin{array}{cc} 1 & 0 \\ 0 & r^2 \end{array} \right)$$
• So $$d \vec{\ell} \cdot d \vec{\ell} = dr^2 + r^2 d \theta^2$$ but $$\left( \vec{d} \phi \right)^r = \frac{\partial \phi}{\partial r} \quad \mathrm{and} \quad \left( \vec{d} \phi \right)^{\theta} = \frac{1}{r^2} \frac{\partial \phi}{\partial \theta}$$
• Often you can guess the correct form just by looking at the relevant units

## Transforming Derivatives in Curved Spaces

• For boosts in SR, transforming vectors was much simpler
• In a curved space, we must transform both the components of the vector, and the basis vectors which we are using
• Think of $\vec{V} = V^{\alpha} \vec{e}_{\alpha}$, we must transform both parts of the RHS
• To transform the derivative: $$\frac{\partial \vec{V}}{\partial x^{\beta}} = \frac{\partial V^{\alpha}}{\partial x^{\beta}} \vec{e}_{\alpha} + V^{\alpha} \frac{\partial \vec{e}_{\alpha}}{\partial x^{\beta}}$$
• We define a new set of objects, the Christoffel symbols (not tensors), which transform basis vectors $$\frac{\partial \vec{e}_{\alpha}}{\partial x^{\beta}} = {\Gamma^{\mu}}_{\alpha \beta} \vec{e}_{\mu}$$
• This allows a rewriting $$\frac{\partial \vec{V}}{\partial x^{\beta}} = \frac{\partial V^{\alpha}}{\partial x^{\beta}} \vec{e}_{\alpha} + V^{\alpha} {\Gamma^{\mu}}_{\alpha \beta} \vec{e}_{\mu}$$
• In Cartesian coordinates, the Christoffel symbols vanish

## Example: Polar coordinates

• Remember $$\frac{\partial \vec{e}_{\alpha}}{\partial x^{\beta}} = {\Gamma^{\mu}}_{\alpha \beta} \vec{e}_{\mu}$$
• And for polar coordinates: $$\frac{\partial}{\partial r} \vec{e}_r = 0$$ $$\frac{\partial}{\partial \theta} \vec{e}_r = \frac{1}{r} \vec{e}_{\theta}$$ $$\frac{\partial}{\partial r} \vec{e}_{\theta} = \frac{1}{r} \vec{e}_{\theta}$$ $$\frac{\partial}{\partial \theta} \vec{e}_{\theta} = -r \vec{e}_{r}$$
• Which allows one to compute, e.g. $${\Gamma^{r}}_{rr}, {\Gamma^{r}}_{r\theta}, {\Gamma^{r}}_{\theta r}, {\Gamma^{r}}_{\theta \theta}, {\Gamma^{\theta}}_{rr}, \ldots$$

## Covariant Derivatives

• By relabeling $$\frac{\partial \vec{V}}{\partial x^{\beta}} = \frac{\partial V^{\alpha}}{\partial x^{\beta}} \vec{e}_{\alpha} + V^{\mu} {\Gamma^{\alpha}}_{\mu \beta} \vec{e}_{\alpha}$$ we make it clear that we can rewrite this as $${V^{\alpha}}_{; \beta} = {V^{\alpha}}_{,\beta} + V^{\mu} {\Gamma^{\alpha}}_{\mu \beta}$$ by defining a new notation with ";"
• This allows us to define the covariant derivative $${\left( \nabla \vec{V} \right)^{\alpha}}_{\beta} = \left( \nabla_{\beta} \vec{V} \right)^{\alpha} = {V^{\alpha}}_{; \beta}$$ which is a $\left( \begin{array}{c} 1 \\ 1 \end{array} \right)$ tensor

## Covariant Derivatives of one-forms

• Schutz shows that $$\left( \nabla_{\beta} \tilde{p} \right)_{\alpha} = \left( \nabla \tilde{p} \right)_{\alpha \beta} \equiv p_{\alpha,\beta} + p_{\mu} {\Gamma^{\mu}}_{\alpha \beta}$$
• Thus there is a product rule for covariant differentiation $$\nabla_{\beta} \left( p_{\alpha} V^{\alpha} \right) = p_{\alpha; \beta} V^{\alpha} + p_{\alpha} {V^{\alpha}}_{; \beta}$$
• And you can also take covariant derivatives of tensors

## Group Work

• Construct a new coordinate system, $\eta(x,y)$ and $\xi(x,y)$ with a non-trivial metric in two dimensions and compute the non-zero Christoffel symbols following Schutz Ch. 5. Compute the transformations, e.g. ${\Lambda^x}_{\eta}$ and use those to determine how the basis vectors transform. Take the derivatives of those basis vectors and then determine the Chistoffel symbols.